If a, b, c are in AP, x is the GM between a and b; y is the GM between b and c; then show that b2 is the AM between x2 and y2.
To prove: b2 is the AM between x2 and y2.
Given: (i) a, b, c are in AP
(ii) x is the GM between a and b
(iii) y is the GM between b and c
Formula used: (i) Arithmetic mean between ![]()
(ii) Geometric mean between ![]()
As a, b, c are in A.P.
⇒ 2b = a + c … (i)
As x is the GM between a and b
⇒ x = ![]()
⇒ x2 = ab … (ii)
As y is the GM between b and c
⇒ y = ![]()
⇒ y2 = bc … (iii)
Arithmetic mean of x2 and y2 is ![]()
Substituting the value from (ii) and (iii)

![]()
Substituting the value from eqn. (i)
![]()
= b2
Hence Proved
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